Solution 4.2:6
From Förberedande kurs i matematik 1
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- | { | + | We can work out the length we are looking for by taking the difference |
- | < | + | <math>a-b\text{ }</math> |
- | + | of the sides | |
+ | <math>a</math> | ||
+ | and | ||
+ | <math>b</math> | ||
+ | in the triangles below: | ||
+ | |||
[[Image:4_2_6_13.gif|center]] | [[Image:4_2_6_13.gif|center]] | ||
[[Image:4_2_6_2.gif|center]] | [[Image:4_2_6_2.gif|center]] | ||
+ | |||
+ | If we take the tangent of the given angle in each triangle, we easily obtain | ||
+ | <math>a</math> | ||
+ | and | ||
+ | <math>b</math>: | ||
+ | |||
+ | |||
[[Image:4_2_6_13.gif|center]] | [[Image:4_2_6_13.gif|center]] | ||
[[Image:4_2_6_4.gif|center]] | [[Image:4_2_6_4.gif|center]] | ||
+ | |||
+ | <math>a=1\centerdot \tan 60^{\circ }=\frac{\sin 60^{\circ }}{\cos 60^{\circ }}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}</math> |
Revision as of 08:29, 29 September 2008
We can work out the length we are looking for by taking the difference \displaystyle a-b\text{ } of the sides \displaystyle a and \displaystyle b in the triangles below:
If we take the tangent of the given angle in each triangle, we easily obtain \displaystyle a and \displaystyle b:
\displaystyle a=1\centerdot \tan 60^{\circ }=\frac{\sin 60^{\circ }}{\cos 60^{\circ }}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}